2015年考研数二第11题解析

题目

设函数 $f(x)$ 连续,$\varphi (x) = \int_{0}^{x^{2}} x f(t)dt$. 若 $\varphi (1) = 1$, $\varphi^{‘} (1) = 5$, 则 $f(1)=?$

解析

已知:

$$
\varphi (x) = x \int_{0}^{x^{2}} f(t)dt.
$$

$$
\varphi^{‘}(x) = \int_{0}^{x^{2}} f(t)dt + x \cdot f(x^{2}) \cdot 2x.
$$

且:

$$
\varphi (1) = \int_{0}^{1}f(t)dt = 1. ①
$$

$$
\varphi^{‘}(1) = \int_{0}^{1}f(t)dt + 2f(1) = 5. ②
$$

联立 $①$, $②$ 两式可知:

$$
1+2f(1) = 5 \Rightarrow
$$

$$
2f(1) = 4 \Rightarrow
$$

$$
f(1) = 2.
$$

综上可知,正确答案为 $2$.

EOF